Problem: There are some tasks which have the following structure: you are given a model, and you can do some operations, you should use these operations to achive the goal. One way to create a new task is to use the same model and same operations, but change the goal.

Let's have a try. I have created the following task for Topcoder SRM 557 Div1-Hard: you are given n integers x1, x2, ..., xn. You are allowed to perform the assignments (as many as you want) of the following form xi ^= xj (in the original task i and j must be different, but in this task we allow i to equal j). The goal is to maximize the sum of all xi.

Now we just change the goal. You are also given n integers y1, y2, ..., yn. You should make x1, x2, ..., xn exactly equal to y1, y2, ..., yn. In other words, for each i number xi should be equal to yi.

Input

The first line contains an integer n (1 ≤ n ≤ 10000). The second line contains n integers: x1 to xn (0 ≤ xi ≤ 109). The third line contains n integers: y1 to yn (0 ≤ yi ≤ 109).

Output

If there is no solution, output -1.

If there is a solution, then in the first line output an integer m (0 ≤ m ≤ 1000000) – the number of assignments you need to perform. Then print m lines, each line should contain two integers i and j (1 ≤ i, j ≤ n), which denote assignment xi ^= xj.

If there are multiple solutions you can print any of them. We can prove that under these constraints if there exists a solution then there always exists a solution with no more than 106 operations.

Examples

Input

2
3 5
6 0


Output

2
1 2
2 2


Input

5
0 0 0 0 0
1 2 3 4 5


Output

-1


Input

3
4 5 6
1 2 3


Output

5
3 1
1 2
2 2
2 3
3 1


Input

3
1 2 3
4 5 6


Output

-1

Note

Assignment a ^= b denotes assignment a = a ^ b, where operation "^" is bitwise XOR of two integers.

Solution (in CPP):

Solution:
#include <bits/stdc++.h>
#pragma comment(linker, "/STACK:36777216")
using namespace std;
const int N = 1e4 + 7;
const int INF = 0x7fffffff;
const int MOD = 1e9 + 7;
const double EPS = 1e-8;
void Debug(int a[], int n) {
  for (int i = 1; i <= n; i++) {
    for (int j = 30; j >= 0; j--) printf("%d", a[i] >> j & 1);
    puts("");
  }
  puts("");
}
int pos[50];
void Gao(int a[], int n, vector<pair<int, int> > &vec) {
  int r = 1;
  for (int i = 30; i >= 0; i--) {
    if ((a[r] >> i & 1) == 0) {
      int pos = -1;
      for (int j = r; j <= n; j++)
        if (a[j] >> i & 1) {
          pos = j;
          break;
        }
      if (pos == -1) continue;
      swap(a[r], a[pos]);
      vec.push_back(make_pair(r, pos));
      vec.push_back(make_pair(pos, r));
      vec.push_back(make_pair(r, pos));
    }
    for (int j = 1; j <= n; j++)
      if ((a[j] >> i & 1) && j != r) {
        a[j] ^= a[r];
        vec.push_back(make_pair(j, r));
      }
    pos[r] = i;
    r++;
  }
}
int Lcp(int a, int b) {
  int tot = 0;
  for (int i = 30; i >= 0; i--) {
    if ((a >> i & 1) != (b >> i & 1)) break;
    tot++;
  }
  return tot;
}
bool Check(int a[], int b[], int n, vector<pair<int, int> > &vec) {
  int r = 1;
  for (int i = 1; i <= 30; i++) {
    if (a[i] == b[i]) continue;
    for (int j = i; j <= 30; j++) {
      if (Lcp(a[i] ^ a[j], b[i]) > Lcp(a[i], b[i]))
        a[i] ^= a[j], vec.push_back(make_pair(i, j));
    }
    if (a[i] != b[i]) return false;
  }
  return true;
}
int n, a[N], b[N];
vector<pair<int, int> > vec1, vec2;
int main() {
  cin >> n;
  for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
  for (int i = 1; i <= n; i++) scanf("%d", &b[i]);
  vector<pair<int, int> > vec1, vec2;
  Gao(a, n, vec1);
  Gao(b, n, vec2);
  if (!Check(a, b, n, vec1)) {
    puts("-1");
  } else {
    for (int i = vec2.size() - 1; i >= 0; i--) vec1.push_back(vec2[i]);
    printf("%d\n", vec1.size());
    for (int i = 0; i < vec1.size(); i++)
      printf("%d %d\n", vec1[i].first, vec1[i].second);
  }
}